Y=(8x-4+12)(3x-8)

Solution for Y=(8x-4+12)(3x-8) equation:

=(8Y-4+12)(3Y-8)

We move all terms to the left:

-((8Y-4+12)(3Y-8))=0

We add all the numbers together, and all the variables

-((8Y+8)(3Y-8))=0

We multiply parentheses ..

-((+24Y^2-64Y+24Y-64))=0


We calculate terms in parentheses: -((+24Y^2-64Y+24Y-64)), so:

(+24Y^2-64Y+24Y-64)

We get rid of parentheses

24Y^2-64Y+24Y-64

We add all the numbers together, and all the variables

24Y^2-40Y-64

Back to the equation:

-(24Y^2-40Y-64)


We get rid of parentheses

-24Y^2+40Y+64=0

a = -24; b = 40; c = +64;
Δ = b2-4ac
Δ = 402-4·(-24)·64
Δ = 7744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7744}=88$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-88}{2*-24}=\frac{-128}{-48}
=2+2/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+88}{2*-24}=\frac{48}{-48}
=-1
$